If $$x + \frac{1}{x} = 3,$$   x ≠ 0 then the value of $${x^7} + \frac{1}{{{x^7}}}$$  is

Solution (By Examveda Team)

$$\eqalign{ & x + \frac{1}{x} = 3,\,x \ne 0 \cr & {x^2} + \frac{1}{{{x^2}}} = {3^2} - 2 = 7 \cr & {x^4} + \frac{1}{{{x^4}}} = {7^2} - 2 = 47 \cr & {x^3} + \frac{1}{{{x^3}}} = {3^3} - 3 \times 3 \cr & = 27 - 9 \cr & = 18 \cr & {x^7} + \frac{1}{{{x^7}}} = \left( {{x^4} + \frac{1}{{{x^4}}}} \right)\left( {{x^3} + \frac{1}{{{x^3}}}} \right) - \left( {x + \frac{1}{x}} \right) \cr & = 47 \times 18 - 3 \cr & = 846 - 3 \cr & = 843 \cr} $$